February 2018 - Page 4 of 7 - Shawn Zhong

Math 541 – 2/14

$Homomorphism • Definition ○ Let G,H be groups ○ A function f:G→H is a homomorphism if ○ f(g_1 g_2 )=f(g_1 )f(g_2 ), ∀g_1,g_2∈G ○ One says f "respects", or "preserves" the group operation • Trivial Examples ○ Let G be a group ○ The identity map of G→G is a homomorphism § f(g_1 )f(g_2 )=1⋅1=1=f(g_1 g_2 ) ○ The map G→G given by g↦1 is a homomorphism § This only works if we send every element of G to 1 § If x∈G∖{1}, and f:G→G is given by g↦x,∀g § f(g_1 g_2 )=f(g_1 )(g_2 ) § i.e. x=x^2 § Thus x=1 § This is impossible since x∈G∖{1} • Example 1 ○ Let f:R→R∗ be given by f(x)=e^x ○ f is a homomorphism ○ f(x_1+x_2 )=e^(x_1+x_2 )=e^(x_1 ) e^(x_2 )=f(x_1 )f(x_2 ) • Example 2 ○ Let G be a group, and let x∈G ○ The map f:G→G, g↦xgx^(−1) is a homomorphism ○ f(g_1 g_2 )=xg_1 g_2 x^(−1)=xg_(1 ) x^(−1) xg_2 x^(−1)=f(g_1 )f(g_2 ) ○ This homomorphism is called conjugation by x • Example 3 ○ Let n∈Z ○ Is f:Z→Z, x↦x+n a homomorphism? ○ Answer: Only when n=0 ○ f(0+0)=f(0) ○ i.e. n+n=n ○ Thus n=0 • Example 4 ○ Let n∈{0,1,2,3,…} ○ Is α:Z→Z, x↦x^n a homomorphism? ○ Answer: Only when n=1 ○ When x=0 § α(x)=1 § 1 is not the identity (0 is) § So this doesn’t work ○ For n≥2 § α(x_1+x_2 )=α(x_1 )+α(x_2 ) § (x_1+x_2 )^n=x_1^n+x_2^n § This is not always true § For instance, when x_1=x_2=1, 2^n≠2 for n≥2 • Example 5 ○ Let n∈Z ○ Is β:Z→Z, x↦nx a homomorphism? ○ Answer: Yes ○ β(x_1+x_2 )=n(x_1+x_2 )=nx_1+nx_2=β(x_1 )+β(x_2 ) • Example 6 ○ The previous examples is a special case of the following: ○ Let G be a group, and n∈Z ○ Define β:G→G, g↦g^n ○ Then β is homomorphism ∀n∈Z iff G is abelian ○ Proof: homomorphism ⇒ abelian § Say n=−1 § Let g_1,g_2∈G § β(g_1,g_2 )=β(g_1 )β(g_2 ) § (g_1 g_2 )^(−1)=g_1^(−1) g_2^(−1) § g_2^(−1) g_1^(−1)=g_1^(−1) g_2^(−1) § (g_2^(−1) g_1^(−1) )^(−1)=(g_1^(−1) g_2^(−1) )^(−1) § (g_1^(−1) )^(−1) (g_2^(−1) )^(−1)=(g_2^(−1) )^(−1) (g_1^(−1) )^(−1) § g_1 g_2=g_2 g_1 § Thus G is abelian ○ Proof: abelian ⇒ homomorphism § (Homework) Isomorphism • Definition ○ Let G,H be groups ○ A homomorphism α:G→H is a isomorphism if ○ there is a homomorphism β:H→G s.t. ○ αβ=id_H and βα=id_G ○ In this case, we say G and H are isomorphic • Fact ○ L:G→H is an isomorphism iff it is a bijective homomorphism ○ Proof: isomorphism ⇒ bijective homomorphism § Clear ○ Proof: bijective homomorphism ⇒ isomorphism § Need to show α^(−1) is a homormorphism § (Homework) • Example ○ R(0)≔{r∈Rr0} is a group under multiplication ○ Define f:R→R(0) where f(x)=e^x ○ Then f is a homomorphism ○ Moreover, f is an isomorphism ○ The inverse of f is ln • Observation ○ If G,H are isomorphic groups, then |G|=|H| Proposition 16 • Statement ○ If f:G→H is an isomorphism, then ○ G is abelian iff H is abelian ○ ∀g∈G, |g|=|f(g)| ○ Note |g|=inf⁡{n∈Z(0)│g^n=1}$

Math 521 – 2/14

$Finite and Countable • Definition ○ Let J_n={1,2,3,…,n} and N={1,2,3,…} ○ For any set A, we say ○ A is finite if A~J_n for some n (∅~J_0 so ∅ is finite) ○ A is infinite if A≁J_n for all n ○ A is countable if A~N ○ A is uncountable if A is neither finite nor countable ○ A is at most coutable if A is finite or countable • Examples ○ N is clearly countable § N={1,2,3,…} ○ Z is countable § Z={0,1,−1,2,−2,3,−3,…} § Define f:N→Z by § f(n)≔{■8(n/2&n is even@(1−n)/2&n is odd)┤ § f is injective □ If f(n)=f(m) □ then n/2=m/2 or (1−n)/2=(1−m)/2 □ Either way, n=m § f is surjective □ Given k∈Z, □ If k0, k=f(2k) □ If k≤0, k=f(−2k+1) § Thus f is bijective ○ Q is countable § There are "less" rational numbers q=m/n (m,n∈Z,n≠0) than § there are ordered pairs of integers (m,n) □ 1/2=15/30 but (1,2)≠(15,30) □ We can also ignore negatives and zeros □ because integers are in 1-1 correspondence with N § Idea: Write ordered pairs of integers in a 2 dimension array § Putting this all together, we have § Q={0,±1,±2,±1/2,±1/3,±3,±4,±3/2,±2/3,±1/4,…} Sequence • Definition ○ A sequence is a function defined on N ○ Notationally, this is often written {x_n } ○ Meaning f(x)=x_n for all n∈N • Example ○ {1/n}={1, 1/2,1/3,…} Theorem 2.8 • Statement ○ Every infinite subset of a countable set is countable • Proof ○ Suppose E⊂A, A is countable and E is infinite ○ Since A is countable, its element will be a sequqnce ○ (order given by the bijective function f:N→A) ○ So, A={x_n } as a set ○ Let n_1 be the smallest n∈N such that x_(n_1 )∈E ○ Let n_2 be the next smallest n∈N such that 〖x_n〗_2∈E ○ So E={x_(n_k ) }={x_(n_1 ),x_(n_2 ),x_(n_3 ),…} (A sequence indexed by k∈N) ○ Now consider g:N→E given by g(k)=x_(n_k ) ○ g is clearly one-to-one and onto by construction ○ So E is countable • Example ○ If A={1, 1/2,1/3,…} and E={1, 1/4,1/9,1/16} ○ Then A={1/n}={1, 1/2,1/3,…} ○ and E={1/n_k }={1, 1/4,1/9,1/16} where n_k=k^2 for k∈N ○ f:A→E by f(k)=1/k^2$

2.4 Sequences and Summations

$Introduction • Sequences are ordered lists of elements. ○ 1,2,3,5,8 ○ 1,3,9,27,81,… • Sequences arise throughout mathematics, computer science, and in many other disciplines, ranging from botany to music. • We will introduce the terminology to represent sequences and sums of the terms in the sequences. Sequences • Definition ○ A sequence is a function from a subset of the integers to a set S. ○ The notation a_n is used to denote the image of the integer n. ○ We can think of a_n as the equivalent of f(n) ○ where f is a function from {0,1,2,…..} to S. ○ We call a_n a term of the sequence. • Example ○ Consider the sequence {a_n } where ○ a_n=1/n, {a_n }={a_1,a_2,a_3,…} ○ 1, 1/2,1/3,… Strings • A string is a finite sequence of characters from a finite set (an alphabet). • Sequences of characters or bits are important in computer science. • The empty string is represented by λ. • The string abcde has length 5. Geometric Progression • Definition ○ A geometric progression is a sequence of the form: a,ar,ar^2,…,ar^n,… ○ where the initial term a and the common ratio r are real numbers • Example 1 ○ Let a = 1 and r = −1. Then: ○ {b_n }={b_0,b_1,b_2,b_3,b_4,…}={1,−1,1,−1,1,…} • Example 2 ○ Let a = 2 and r = 5. Then: ○ {c_n }={c_0,c_1,c_2,c_3,c_4,…}={2,10,50,250,1250,…} • Example 3 ○ Let a = 6 and r=1/3. Then: ○ {d_n }={d_0,d_1,d_2,d_3,d_4,…}={6,2, 2/3,2/9,2/27,…} Arithmetic Progression • Definition ○ A arithmetic progression is a sequence of the form ○ a,a+d,a+2d,…,a+nd,… ○ where the initial term a and the common difference d are real numbers • Example 1 ○ Let a = −1 and d = 4: ○ {s_n }={s_0,s_1,s_2,s_3,s_4…}={−1,3,7,11,15…} • Example 2 ○ Let a = 7 and d = −3: ○ {t_n }={t_0,t_1,t_2,t_3,t_4…}={7,4,1,−2,−5…} • Example 3 ○ Let a = 1 and d = 2: ○ {u_n }={u_0,u_1,u_2,u_3,u_4…}={1,3,5,7,9…} Recurrence Relations • A recurrence relation for the sequence {a_n} is an equation that ○ expresses a_n in terms of a_0, a_1, …, a_(n−1) ○ for all integers n with n≥n_0, where n_0 is a nonnegative integer. • A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation. • The initial conditions for a sequence specify the terms that precede the first term where the recurrence relation takes effect. • Example ○ Let {a_n} be a sequence that satisfies § the recurrence relation a_n=a_(n−1)+3 for n=1,2,3,4 § and suppose that a_0=2. ○ What are a_1, a_2 and a_3? ○ a_1=a_0+3=2+3=5 ○ a_2=5+3=8 ○ a_3=8+3=11 Fibonacci Sequence • Define the Fibonacci sequence, f0 ,f1 ,f2,…, by: ○ Initial Conditions: f_0=0, f_1=1 ○ Recurrence Relation: f_n=f_(n−1)+f_(n−2) • Find f_2,f_3,f_4,f_5,f_6 ○ f_2=f_1+f_0=1+0=1 ○ f_3=f_2+f_1=1+1=2 ○ f_4=f_3+f_2=2+1=3 ○ f_5=f_4+f_3=3+2=5 ○ f_6=f_5+f_4=5+3=8 Solving Recurrence Relations • Finding a formula for the n-th term of the sequence generated by a recurrence relation is called solving the recurrence relation. • Such a formula is called a closed formula. • Various methods for solving recurrence relations will be covered in Chapter 8 where recurrence relations will be studied in greater depth. • Here we illustrate by example the method of iteration in which we need to guess the formula. The guess can be proved correct by the method of induction (Chapter 5). • Method 1: Working upward, forward substitution ○ Let {a_n} be a sequence that satisfies the recurrence relation ○ a_n=a_(n−1)+3 for n = 2,3,4,… and suppose that a_1=2 ○ a_2=2+3 ○ a_3=(2+3)+3=2+3∙2 ○ a_4=(2+2∙3)+3=2+3∙3 ○ ⋮ ○ a_n=a_(n−1)+3=(2+3∙(n–2))+3=2+3(n–1) • Method 2: Working downward, backward substitution ○ Let {a_n} be a sequence that satisfies the recurrence relation ○ a_n=a_(n−1)+3 for n = 2,3,4,… and suppose that a_1=2 ○ a_n=a_(n−1)+3 ○ =(a_(n−2)+3)+3=a_(n−2)+3∙2 ○ =(a_(n−3)+3)+3∙2=a_(n−3)+3∙3 ○ =… ○ =a_2+3(n–2)=(a_1+3)+3(n–2)=2+3(n–1) Summations • Sum of the terms a_m,a_(m+1),…,a_n from the sequence {a_n } • Notation for a_m+a_(m+1)+…+a_n ○ ∑_(j=m)^n▒a_j • The variable j is called the index of summation. It runs through all the integers starting with its lower limit m and ending with its upper limit n. • More generally for a set S ○ ∑_(j∈S)▒a_j • Examples: ○ r^0+r^1+r^2+r^3+…+r^n=∑_(j=0)^n▒r^j ○ 1+1/2+1/3+1/4+…=∑_(i=1)^∞▒1/i ○ If S={2,5,7,10} then ∑_(j∈S)▒a_j =a_2+a_5+a_7+a_10 Geometric Series • Sums of terms of geometric progressions • Product Notation • Sum of the terms a_m,a_(m+1),…,a_n from the sequence {a_n } • Notation for a_m×a_(m+1)×…×a_n ○ ∏_(j=m)^n▒a_j$

Math 521 – 2/12

$Schwarz Inequality • See Theorem 1.35 in Rudin for a proof of Schwarz Inequality for ℂ ○ For intuition, try proving (x_1 y_2+x_2 y_2 )^2≤(x_1^2+x_2^2 )(y_1^2+y_2^2 ) • Triangle Inequality ○ In a Euclidean Space, |x ⃗⋅y ⃗ |≥|x ⃗ |⋅|y ⃗ | ○ |x ⃗+y ⃗^2 |=|x ⃗ |^2+2x ⃗⋅y ⃗+|y ⃗ |^2≤|x ⃗ |^2+2|x ⃗ ||y ⃗ |+|y ⃗ |^2=(|x ⃗ |+|y ⃗ |)^2 ○ Thus |x ⃗+y ⃗ ||x ⃗ |+|y ⃗ | ○ Let x ⃗≔x ⃗−y ⃗, y ⃗≔y ⃗−z ⃗, we have |x ⃗−z ⃗ ||x ⃗−y ⃗ |+|y ⃗−z ⃗ | Function • Given two sets A and B • A function (or mapping) is a rule that assigns elements in A to elements in B • Notationally, if f is a function from A to B, we write f:A→B • Set A is called the domain of f • Set B is called the codomain of f • For E⊂A, f(E)={b∈B│b=f(e) for some e∈E} is the image of E under f • f(A) is called the range of f • If f(A)=B, then we say that f is onto or surjective • If f(a_1 )=f(a_2 ) implies a_1=a_2, then f is one-to-one or injective • A function that is both one-to-one and onto is said to be bijective • For E⊂B, f^(−1) (E)={a∈A│f(a)∈E} is the inverse image of E under f • Notationally, if y∈B, f^(−1) (y)=f^(−1) ({y}) ○ f^(−1) is at most a single element set for all y∈B if and only if f is injective ○ In this case, f^(−1) can be thought of as a function maps to the single element • Example ○ f:R→R defined by f(x)=x^2 ○ f^(−1) ({1})={1,−1} ○ f^(−1) ({x∈Rx0})=∅ ○ f^(−1) ({0})={0}, we can also write f^(−1) (0)=0 Cardinality • If there exists a one-to-one, onto mapping from set A to set B • We say that A and B can be put in one-to-one correspondence • And that A and B have the same cardinality (or cardinal number) • In this case, we write A~B Equivalence Relation • One-to-one correspondence is an example of an equivalence relation • An equivalence relation satisfies 3 properties ○ Reflexive: A~A ○ Symmetric: If A~B, then B~A ○ Transitivity: If A~B, B~C, then A~C Finite and Countable • Let J_n={1,2,3,…,n} and N={1,2,3,…} • For any set A, we say ○ A is finite if A~J_n for some n (∅~J_0 so ∅ is finite) ○ A is infinite if A≁J_n for all n ○ (To be continued)$

Math 541 – 2/12

$Proposition 15 • Statement ○ |S_n |=n! • Proof ○ First, we prove that if X and Y are sets of order n, then ○ There are n! injective functions from X to Y ○ We argue by induction on n ○ n=1 § Clear ○ n 1 § Suppose f:X→Y is injective § Let x∈X, then there are n possibilities for f(x) § f restricts to an injective function X∖{x}→Y∖{f(x)} § There are (n−1)! such functions, by induction § Thus, there are n(n−1)!=n! injective functions X→Y ○ Now, take X={1,…,n}=Y ○ Then every injection between finite sets of the same order is a bijection ○ Notice that the sets must be finite (Counterexample: f:Z→Z, n↦2n) Cycle • Definition ○ Fix n∈Z( 0). Let a_1,…,a_t∈{1,…,n} ○ The element of S_n given by § a_i↦a_(i+1) for 1≤i≤t−1 § a_t↦a_1 § j↦j if j∉{a_1,…a_t } ○ is denoted by (a_1,a_2,…,a_t ) and is called a cycle of length t • Example ○ Let σ=(1 3 2)∈S_4, then ○ (■8(i&1&2&3&4@↓&↓&↓&↓&↓@σ(i)&3&1&2&4)) ○ Notice: (1 3 2)=(3 2 1)=(2 1 3) Disjoint Cycles • Definition ○ Two cycles (a_1,…a_t ) and (b_1,…,b_k ) are disjoint if ○ {a_1,…a_t }∩{b_1,…,b_k }=∅ • Example ○ (1 2), (3 4)∈S_4 are disjoint • Fact ○ Every element of S_n can be written as a product of disjoint cycles ○ S_1={(1)} ○ S_2={(1),(1 2)} ○ S_3={(1), (1 2), (1 3), (2 3),(1 2 3), (1 3 2)} ○ S_4 = {(1), (1 2), (1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3), (1 2 4), (1 3 2), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2), (1 2)(3 4), (1 4)(2 3), (1 3)(2 4s)} ○ Note: We write the identity of S_n as (1) Cycle Decomposition for Permutations • Algorithm Method Example a≔min⁡{x∈Nx not appeared in previous cycles} (1 Begin the new cycle: (a b≔σ(a) σ(1)=12=b If b=a 12≠1 • close the cycle with a right parenthesis So write (1 12 • return to step 1 If b≠a • write b next to a in this cycle: (a b c≔σ(b) σ(12)=8 If c=a 8≠1 • close the cycle with a right parenthesis So continue the cycle as: • return to step 1 (1 12 8 If c≠a • write c next to in this cycle: (a b c b≔c and repeat this step until the cycle closes Naturally this process stops when all the numbers from {1,2, …,n} have appeared in some cycle. σ = (1 1 2 8 10 4)(2 1 3)(3)(5 1 1 7)(6 9) Remove all cycles of length 1 σ = (1 1 2 8 10 4)(2 1 3)(5 1 1 7)(6 9) • Example ○ Take σ∈S_13 to be the following ○ (■8(i&1&2&3&4&5&6&7&8&9&10&11&12&13@↓&↓&↓&↓&↓&↓&↓&↓&↓&↓&↓&↓&↓&↓@σ(i)&12&13&3&1&11&9&5&10&6&4&7&8&3)) ○ Start with 1, σ(1)=12, so write 12 after 1. ○ Keep going until you cycle back to 1 ○ Start with the smallest number which hasn t yet appeared, and repeat. ○ Repeat this step until 1,…,13 have all appeared. Product of Cycles • Reminder: Read from right to left • Example ○ Write σ=(1 2 3)(1 2)(3 4) as a product of disjoint cycles ○ What is σ(1)? § (3 4) maps 1 to 1 § (1 2) maps 1 to 2 § (1 2 3) maps 2 to 3 § Thus σ(1)=3 ○ Similarly σ(3)=4, σ(4)=1 ○ Thus we close the cycle (1 3 4) ○ We won t write down (2), since it is the identity ○ Thus σ=(1 3 4)(2)=(1 3 4) ○ Note: σ∈S_4, but it make sense to think of σ∈S_n for n 4 • Commutativity of S_n ○ (1 2)(1 2 3)=(2 3) ○ (1 2 3)(1 2)=(3 1) ○ In particular S_3 is not abelian ○ Therefore S_n is not abelian for n≥3$

Shawn Zhong

Shawn Zhong

Math 541 – 2/14

Math 521 – 2/14

2.4 Sequences and Summations

Math 521 – 2/12

Math 541 – 2/12

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